# Moniod with a unit

• Oct 21st 2009, 05:45 PM
sfspitfire23
Moniod with a unit
Hello!

If M is a finite monoid and $au=bu$ in M implies that $a=b$, show that $u$ is a unit.

with hints: If $M=\{a_1,\cdots a_n\}$ show that $a_1u,\cdots , a_nu$ are distinct.

Attempt:

If $M=\{a_1,\cdots a_n\}$ then by definition it has an identity element $e$. Then, because M is finite, there will be an element which will map to the identity of M. Each $a_1u\cdots a_nu$ will be distinct as for every $a_n$, u will map it back to itself. Thus, $au=bu$ which, because $u$ will map $a$ to itself $a(e)=b(e)$ and $a=b(e)$ so $u$ must be a unit.

Thanks guys!
• Oct 22nd 2009, 03:50 AM
clic-clac
Hi

What you have to prove is: $\exists b\in M\ \text{s.t.}\ bu=e.$

To show the hint, just use that $au=bu\Rightarrow a=b,$ (for instance prove it by contraposition).

Assume you've proved it. Show that $M=\{a_1u,...,a_nu\}$ (using a property of finite sets)

Therefore $e\in M$ means $e\in \{a_1u,...,a_nu\}$. Can you conclude?
• Oct 22nd 2009, 09:10 PM
sfspitfire23
would it suffice to create a cayley table with elements 1,a,b and show that a and b both have an inverse? ugh, where would i go from there?

thanks, really appreciate it
• Oct 22nd 2009, 11:12 PM
clic-clac
I don't really understand, why just $e,a,b$ in a Cayley table. $ab$ can be different from $e,a$ or $b.$

What the proof says is : the condition you have implies that the map $M\rightarrow M: a\mapsto au$ is injective and then surjective (because it's an injective map between sets of same finite cardinality) therefore $e$ has an antecedant, which is some $b\in M$ such that $bu=e,$ and we're done.